How would I count the days from a date till the first of the following month
Example:
--Start Date
07-07-2011
How many days till:
-- The 1st of the succeeding month of the start date above
08-01-2011
Expected Result (in days):
25
So if I counted the day I get 25, so running this query gets me the desired timestamp:
SELECT CURRENT_DATE + INTERVAL '25 DAYS'
Results:
2011-08-01 00:00:00
just can't think of a way to get the number of days, any suggestions?
Or start date, end date, number of days between?
I don't have a PostgreSQL server handy, so this is untested, but I would try:
SELECT (DATE_TRUNC('month', CURRENT_DATE) + INTERVAL '1 MONTH') - CURRENT_DATE
Related
I need to return the dates between the 05 of the last month and the 05 of the current month Example today is the 16/08/2022 I recuperate therefore the whole of the days between the 05/07/2022 and the 05/08/2022
For the moment I try with this query
SELECT DATE from Db_name where date between date_trunc('month', current_date-1) and date_trunc('month',current_date)
Your initial query is very much on track. Just a couple additional things about dates:
current_date-1 subtracts 1 day, you need to subtract 1 month. Thus current_date - interval '1 month'.
date_trunc(somedate) returns the 1st of the month so
date_trunc('2022-08-17') returns 2022-08-01.
getting to the 5th of the month from the 1st just add interval '4 days'.
adding or subtracting intervals to dates results in timestamps.
Since you want dates so needs to be cast back to date.
select *
from db_name
where some_date between (date_trunc('month', current_date-interval '1 month') + interval '4 days')::date
and (date_trunc('month', current_date) + interval '4 days')::date;
But I will repeat do not any reserved word or data type as a name. At best is causes confusion, at worst it will run but do the wrong thing.
I used postgresql to solve the quesion, query to return the day of the week of the first day of the month two years from today. I was able to solve it with the query below, but I am not sure my query is correct, I just wanna make sure
select cast(date_trunc('month', current_date + interval '2 years') as date)
You are correctly computing the first day of the month two years later with:
date_trunc('month', current_date + interval '2 years')
If you want the corresponding day of the week, you can use extract();
extract(dow from date_trunc('month', current_date + interval '2 years'))
This gives you an integer value between 0 (Sunday) and 6 (Saturday)
I am trying to query Firebird to get data from last month, from day 1 until last day (30 or 31 depending on the month). When I use the code below it gives me shifted dates from current, for example day 11/14/2017 until 12/13/2017.
The code:
WHERE DATE >= DATEADD(MONTH,-1, CURRENT_TIMESTAMP(2)) AND DATE<= 'TODAY'
The desired output is 11/01/2017 - 11/30/2017
What is the correct way to do it?
I don't use Firebird but I've used PostgreSQL fairly extensively and I think this should work:
WHERE
DATE BETWEEN dateadd(month, -1, CURRENT_DATE - EXTRACT(DAY FROM CURRENT_DATE) + 1)
AND CURRENT_DATE - EXTRACT(DAY FROM CURRENT_DATE)
Explanation
CURRENT_DATE - EXTRACT(DAY FROM CURRENT_DATE) + 1 should go back to the first of this month and dateadd with -1 month should take it to the previous month. Then if you're between CURRENT_DATE - EXTRACT(DAY FROM CURRENT_DATE) or in other words 12/13/2017 - 13 days that should be the last day of November. Crossing my fingers. Good luck.
Say, I have the following table:
CREATE TABLE Data (
Id INTEGER PRIMARY KEY,
Value DECIMAL,
Date DATE);
Since the application is finance-related, user may choose, which day would be the first day of the month. For instance, if he receives salary every 10th of the month, he may set the first day of the month to be 10th.
I'd like to create a query, which returns average value for n-th day of month, as defined by user. For instance:
Date | Value
---------------+------
10.01.2016 | 10
11.01.2016 | 15
10.02.2016 | 20
11.03.2016 | 10
Result of the query should be:
Day | Average
----+--------
1 | 15
2 | 12.5
Note, that if user sets first day to 10th, 9th of the month may be 28th, 29th, 30th or 31st day of a month (depending on which month we're talking about). So this is not as simple as extracting day number from the date.
Assuming that the date values do not use the format dd.mm.yyyy but one of the supported date formats, you can use the built-in date functions to compute this.
To compute the difference, in days, between two dates, convert them into a date format that uses days as a number, i.e., Julian days.
To get the 'base' day for a month, we can use modifiers:
> SELECT julianday('2001-02-11') -
julianday('2001-02-11', 'start of month', '+10 days') + 2;
2.0
(The +2 is needed because we add to the 1st of the month, not the 0th, and we count beginning at 1, not 0.)
If the day is before the tenth, the computed value would become zero or negative, and we have to use the previous month instead:
> SELECT julianday('2001-02-09') -
julianday('2001-02-09', 'start of month', '-1 month', '+10 days') + 2;
31.0
Combining these results in this expression to compute the n for a date Date:
CASE
WHEN julianday(Date) -
julianday(Date, 'start of month', '+10 days') + 2 > 0
THEN julianday(Date) -
julianday(Date, 'start of month', '+10 days') + 2
ELSE julianday(Date) -
julianday(Date, 'start of month', '-1 month', '+10 days') + 2
END
You can the use this in your query:
SELECT CASE...END AS Day,
AVG(Value) AS Average
FROM Data
GROUP BY Day;
Seems like I found in parallel a different solution:
SELECT avg(Amount),
strftime("%d", Date, "-9 days") AS day
FROM (
SELECT sum(Amount) AS Amount,
strftime("%Y-%m-%d", Date, "start of day") AS Date
FROM Operations
GROUP BY strftime("%Y-%m-%d", Date, "start of day")
)
GROUP BY day;
Where "-9 days" is for 10th of the month (so first_day-1).
Need help figuring out how to determine if the date is the same 'day' as today in teradata. IE, today 12/1/15 Tuesday, same day last year was actually 12/2/2014 Tuesday.
I tried using current_date - INTERVAL'1'Year but it returns 12/1/2014.
You can do this with a bit of math if you can convert your current date's "Day of the week" to a number, and the previous year's "Day of the week" to a number.
In order to do this in Teradata your best bet is to utilize the sys_calendar.calendar table. Specifically the day_of_week column. Although there are other ways to do it.
Furthermore, instead of using CURRENT_DATE - INTERVAL '1' YEAR, it's a good idea to use ADD_MONTHS(CURRENT_DATE, -12) since INTERVAL arithmetic will fail on 2012-02-29 and other Feb 29th leap year dates.
So, putting it together you get what you need with:
SELECT
ADD_MONTHS(CURRENT_DATE, -12)
+
(
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = CURRENT_DATE)
-
(SELECT day_of_week FROM sys_calendar.calendar WHERE calendar_date = ADD_MONTHS(CURRENT_DATE, -12))
)
This is basically saying: Take the current dates day of week number (3) and subtract from it last years day of week number (2) to get 1. Add that to last year's date and you'll have the same day of the week as current date.
I tested this for all dates between 01/01/2010 and CURRENT_DATE and it worked as expected.
Why don't you simply subtract 52 weeks?
current_date - 364
The SQL below will get you to the abbreviated name for the day of week, it's cumbersome but it works across versions of Teradata.
SELECT CAST(CAST(ADD_MONTHS(CURRENT_DATE, -12) AS DATE FORMAT 'E3') AS CHAR(3)) AS LY_DayOfWeek
, CAST(CAST(CURRENT_DATE) AS DATE FORMAT 'E3') AS CHAR(3)) AS CY_DayOfWeek
Dates are internally represented at integers in Teradata as (Year-1900) * 100000 + (MONTH * 100) + DAY. You may be able to do some creative arithmetic to figure out that 12/1/2015 Tuesday was 12/2/2014 Tuesday last year.